Quantum Mechanica Modes of Vibration and Rotation of Moecues Chapter 18 Moecuar Energy Transationa Vibrationa Rotationa Eectronic Moecuar Motions Vibrations of Moecues: Mode approximates moecues to atoms joined by springs. A vibration (one type of a norma mode of vibration) of a CH moiety woud ook ike; Frequency of vibration cassica approach Water (3) http://www.youtube.com/watch?v=1uevvkkw0 CO (4) http://www.youtube.com/watch?v=w5gimzfy6i http://en.wikipedia.org/wiki/moecuar_vibration O (1) http://www.youtube.com/watch?v=5qc4ovadkhs The motions are considered as harmonic osciators. For a moecue of N atoms there are 3N-6 norma modes (noninear) or 3N-5 (inear). 1
During a moecuar vibration the motion of the atoms are with respect to the center of mass, and the center of mass is stationary as far as the vibration is concerned. This concept is true for a norma modes of vibrations of moecues. Working with center of mass coordinates simpifies the soution. Vibrationa motion -harmonic osciator, KE and PE cassica approach Diatomics: Center of mass, does not move. Force constant of spring = k Whether pued apart or pushed together from the equiibrium position, the spring resists the motion by an opposing force. Center of mass coordinates Second Law reduced mass, µ; Soutions (genera) of the DE wi be of the form; Use Euer s Formua µ Spring extension of a mass µ from it s equiibrium position. The physica picture changes from masses (m 1 and m ) connected by a spring (force constant k) to a reduced mass, µ,connected by a spring (samek) to an immovabe wa. where b = c + c and b = ic ( c ) 1 1 1
Ampitudes are rea numbers, b 1 and b are rea or = 0. Appying the BC, at t = 0; x(0) = 0, v(0) = v 0. = 1 and to find T period: 1 Frequency; ν = T Therefore: v = 0 v = v 0 x=0 v = 0 a -a = k anguar veocity = ω = = ν µ where ν = frequency α = phase ange Genera equation; Energy terms (KE, PE) are; 1 1 KE = µ v PE = kx where v and x are veocity and position (dispacement from equiibrium). α = π/ Study Exampe Probem 18.1 α 0 No restriction on E vaues, cassicay. 3
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Harmonic osciator potentia function http://www.youtube.com/watch?v=5qc4ovadkhs Harmonic osciator potentia function Morse curve At room temperature the potentia function very cosey foow the quadratic function. Vibrationa motion is equivaent to a partice of mass µ, vibrating about its equiibrium distance, r 0. x, x Simiar for ow energy situations ~ ground state https://www.youtube.com/watch?v=3rqeir8ntmi How can a partice ike that be described by a set of wave functions. r 0 Our interest is to mode the osciatory motion of the diatomic moecue (reative motions of atoms). Aso of interest kinetic and potentia energy of the diatomic moecue during the osciatory motion. Not necessariy the energies of individua atoms of the moecue. The utimate goa is to find the (eigen) energies and the eigen functions (wavefunction) of the vibrationa states of the diatomic by soving the Schrödinger equation; Recipe to Construct the Schrodinger equation 1. Expression for tota energy (cassica). Construct the tota energy Operator, the Hamitonian, by expressing KE in terms of momentum operator and mass, PE in terms of position operator. 3. Setup the Schrodinger equation, eigenequation H ψ = Eψ. 5
1. Expression for tota vibrationa energy, 1 1 Etot, vib = KE + PE = µ v + kx p 1 = + kx µ. Construct the tota energy Operator, the Hamitonian 1 1 1 d 1 H = p + kx ɵ = i kx µ µ ħ + dx 3. Set the Schrodinger equation, eigenequation H ψ = Eψ Soutions (i.e. normaized wavefunctions, eigenfunctions) of which are; With the normaization constant The first six Hermite poynomias 1 Hn( α / x) = Hermite poynomias n = vibrationa quantum number Hermite poynomias: d Hn( z) = ( 1) e e dz n n z z n 6
The first few wavefunctions (n = 0,..3) The first few wavefunctions, ψ n (n = 0,..4) The respective eigen energies are; zpe Note: the resembance to 1D box non-infinite potentia we resuts. Equa energy gaps are equa in contrast to 1D box case. The first few wavefunctions, ψ n (n = 0,..4) Note the resembance to 1D box. The constraint imposedon the partice by a spring resuts in zero point energy. The energy vaues of vibrationa states are precisey known, but the position of the partices (as described by the ampitude) is imprecise, ony a probabiity density ψ (x)of x can be stated, [ ɵ x, H ] 0. Note: the overfow of the wave function (forbidden) of the wavefunction beyond the potentia barrier wa. 7
At high quantum numbers n(high energy imit) the system gets coser to a cassica system (red -probabiity of x in q.m. osciator, bue - probabiity of x cassica osciator). n = 1 n >>0 high ow high high ow high Bue to back, a partice in a barre cassica mode. Rotationa energies of a cassica rigid rotor (diatomic). In vibrationa motion, veocity, acceeration and momentum are parae to the direction of motion. In the center of mass coordinate system, rigid rotor motion equivaent to a mass of µmoving in a circe of radius r 0 (= bond ength, constant) with an anguar veocity ωand a tangentia veocity v. Rotation in D (on a pane) y No opposing force to rotation no PE (stored of energy) term. A energy = KE. r 0 x r 0 8
p = inear momentum Anguar veocity vector Acceeration in the radia direction Anguar veocity ω and anguar acceeration α, r 0 RHR Mass with constant ω makes α = 0. Tangentia inear veocity, Anguar veocity (~ momentum) vector norma to the pane of motion. (coming out of the pane) ω = E I = I, moment of inertia QM Anguar momentum (D): = r p The rotation on the x,y pane (D) occurs freey. Equivaenty the partice of mass µ= moves on the circe p anguar momentum Magnitude of = (constant radius r = r 0 ) freey; E pot = V(x,y) = 0 φ= 90 o p Etot, rot = KE + PE = KE = r µ Energy Rotationa: p = r Cassica rotor -no restriction on (or E. SE: p ħ H = = + µ µ x y 9
0 < φ < 0 < θ < π r sinθdθdφdr r µ p r In poar coordinates (variabe is φ): BC: Φ ( φ) = Φ ( φ + ) Genera soution for DE), wavefunctions: Rotationa eigenfunction. m = integer cockwise counter-cockwise rotation rotation m = rotationa quantum number, quantization, next page e e BC: Φ ( φ) = Φ ( φ + ) imφ im( φ + ) imφ = e = e = e im i.e. e = 1 im( φ + ) imφ im Now; cos( m ) + isin( m ) = 1 cos( m) = 1 rea number Therefore m= 0, ±1, ±,... e Using the Euer s Formua Note: One boundary condition eads to one q.n., m. 10
Normaization constant A +φ; e e BC: Φ ( φ) = Φ ( φ + ) imφ im( φ + ) imφ = e = e = e im i.e. 1 e im( φ + ) imφ im = Now; cos( m) + isin( m ) = 1 thus cos( m ) = 1 rea number Therefor e m= 0, ±1, ±,... e Normaization condition 0 Φ * ( φ) Φ ( φ) dφ = 1 m m imφ + imφ A+ φ e e d 0 φ = 1 + φ φ A+ φ 0 A d A + φ 1 = ( π ) = 1 = 1 Normaization constant A ±φ; Normaization condition 0 Φ * ( φ) Φ ( φ) dφ = 1 m imφ ± imφ A± φ e e d 0 φ = 1 ± φ φ A± φ 0 A d A ± φ Same regardess of m vaue. = 1 m ( π ) = 1 = 1 1 ± im Φ ± ( φ) = e φ Rotationa wave function m is an integer. rea part 1 + im φ 1 Φ + ( φ) = e = ( cos mφ + i sin mφ) Stationary state - exist 11
1 ± im Φ ± ( φ) = e φ 1 cos φ 1 cos φ 1 cos 3 φ 1 cos 4 φ Non-stationary state - Unacceptabe wavefunction, annihiates 1 cos 5 φ 1 cos 6 φ Energy of rotationa states: Note;No non-zero ZPE. ZPE appears if the potentia energy confinement exists, not here ( V(φ)=0) 1 ± im Φ ± ( φ) = e φ Degenerate states - two fod degenerate (of equa energy) E ω = and anguar momentum = Iω I E ħ m ħm ω = = = I I I I I m = Em m being an integer, rotationa frequency ωwi take finite vaues. discrete set of rotationa frequencies!! For D rotor, = z. 1 E = = = Iω µ r I 0 1
The wavefunctions,. of a rigid rotor are simiar to the wavefunctions of a free partice restricted to a circe (partice in a ring)! For a rigid rotor, anguar momentum ies in the z direction. y z x Anguar momentum z component operator; ɵ z = iħ φ Anguar Momentum D rigid rotor. φ Operator: ɵ z = iħ p r anguar anaogue of momentum 1 ± imφ Φ ± ( φ) = e wave functions Note: for D rigid rotor [ H, ɵ z] = 0 Both operators has the same eigenfunctions. Hence anguar momentum (z component) for this case can be precisey measured. But the position, here represented by φ, cannot be measured precisey; the probabiityfor any interva of dφ is the same, [ ɵɵ φ, z] 0 Anguar momentum (in z direction) is quantized!! Note: for D rigid rotor both H, ɵ have same Φ, [ H, ɵ z z] = 0 13
QM Anguar momentum (3D): Partice on a circe (D). r 0 r r 0 r D 3D Partice on the surface of a sphere (3D). Cassica 3D rotor Partice of mass µon the surface of a sphere of radius r. PE confinement none; V(θ,φ) = 0 Set PE operator set to zero. E = KE + PE = KE KE (energy) operator for rotor in poar coordinates: H rot = Just the steps in soving the SE for rigid rotor a. coect the constants: note b. mutipy by sin θ, rearrange to get SE: Note the form of H rot Differentiation ony w.r.t.; θ φ Wavefunction; Therefore ;separation of variabes possibe. spherica harmonic functions 14
c. substitute for Y(θ,φ) divide by Θ(θ)Φ(φ), we get soutions Because each side of the equation depends ony one of the variabes and the equaity exists for a vaues of the variabes, both sides must be equa to a constant. BC eads to q.n.; BC eads to more q.n.; soutions For a givenq.n. there are (+1), m vaues. That is the state described by a quantum number is comprised of (+1) sub-states of equa energy. Degeneracy of state with q. n., is+1. spherica harmonic functions written in more detai woud take the form: Energy of degenerate states of q. n. : µ r0e I β = = E = ( + 1) ħ ħ E ħ = ( + 1) I quantization of rotationa energy 15
Eigen-energy: Verifiabe in SE: H rot E ħ = ( + 1) I Spherica harmonics are eigen functions of the tota rotationa energy operator. A m states of a 3D rotor has the same energy, E! (degenerate) the quantum number m determines the z-component of the anguar momentum vector. Quantization of Anguar Momentum As wi be seen ater, the shapes, directiona properties and degeneracy of atomic orbitas are dependent on the q.n. and m vaes associated with these orbitas. For 3D roto r; E tot = H = ɵ I I Constant for rotor. Therefore both operators have a common set of eigenfunctions. The operators shoudcommute! [ H, ɵ ] = 0 Now; and therefore Components of the anguar momentum in x, y and z. The respective operators are (Cartesian coordinates): Note that the above operators differ ony by the mutipication constant(1/i), thus the eigen vaues differ by (1/I). Aso note the cacuated (and measurabe) anguar momentum quantity (of precise(eigen) vaue) is that of (therefore ). = ħ ( + 1) H = ɵ I 16
The respective operators are in spherica coordinates woud be: The foowing commutator reationships exist. = f( θ, φ) 0 The component operators do not commute with one another = f( φ) As a resut the direction of vector cannot be specified (known) precisey for rotations in 3D in QM systems as opposed to cassica D rigid rotors. Need to know a vectors at the same time to specify direction of vector. However, anguar momentum wavefunction is an eigenfunction of the Hamitonian too and therefore, Examining z ; H = ɵ I Concusion Spherica harmonics are eigenfunctions of and. z Magnitudes of both and z can be known simutaneousy and precisey, but not the xand the zcomponent vaues of the anguar momentum vector,.!!! ħ z=m Note/ show expicity: ɵ ɵ z [, ] = 0 17
The Picture: Vector mode of anguar momentum H, and commute and, and do not commute. z x y z with one another. Exampe = case. m = + Exampe = case. = ( + 1) ħ = mħ z ħ ħ = ħ 18
Spatia quantization m = + Vector mode of anguar momentum Spherica harmonic wavefunctions and their shapes: = 0 = 1 = Compex functions - visuaization not possibe. 19